Chapter 10, Problem 24P

(a)

To determine

Evaluate ${\cos }^{-1}\left[\overrightarrow{A}\cdot \overrightarrow{B}/AB\right]$.

Answer to Problem 24P

It is found that $\underset{\_}{{\cos }^{-1}\left[\overrightarrow{A}\cdot \overrightarrow{B}/AB\right]=168°}$.

Explanation of Solution

The vector $\text{A}$ is given by.

  $\overrightarrow{A}=-3\widehat{i}+7\widehat{j}-4\widehat{k}$

Vector $\text{B}$ is given as.

  $\overrightarrow{B}=6\widehat{i}-10\widehat{j}+9\widehat{k}$

Take the dot product of vector $\text{A}$, and $\text{B}$.

  $\begin{array}{c}\overrightarrow{A}\cdot \overrightarrow{B}=\left(-3\widehat{i}+7\widehat{j}-4\widehat{k}\right)\cdot \left(6\widehat{i}-10\widehat{j}+9\widehat{k}\right)\\ =-3.00\left(6.00\right)+7.00\left(-10.0\right)+\left(-4.00\right)\left(9.00\right)\\ =-124\end{array}$

Calculate the magnitude of vector $AB$.

  $\begin{array}{c}AB=\sqrt{{\left(-3.00\right)}^2+{\left(7.00\right)}^2+{\left(-4.00\right)}^2}\cdot \sqrt{{\left(6.00\right)}^2+{\left(-10.0\right)}^2+{\left(9.00\right)}^2}\\ =127\end{array}$

Conclusion:

Substitute, $-124$ for $\overrightarrow{A}\cdot \overrightarrow{B}$, and $127$ for $AB$ in ${\cos }^{-1}\left[\overrightarrow{A}\cdot \overrightarrow{B}/AB\right]$.

  $\begin{array}{c}{\cos }^{-1}\left[\overrightarrow{A}\cdot \overrightarrow{B}/AB\right]={\cos }^{-1}\left(\frac{-124}{127}\right)\\ ={\cos }^{-1}\left(-0.979\right)\\ =168°\end{array}$

Therefore, ${\cos }^{-1}\left[\overrightarrow{A}\cdot \overrightarrow{B}/AB\right]$ is equal to $\underset{\_}{168°}$.

(b)

To determine

Evaluate ${\sin }^{-1}\left[\left|\overrightarrow{A}\times \overrightarrow{B}\right|/AB\right]$.

Answer to Problem 24P

It is found that $\underset{\_}{{\sin }^{-1}\left[\left|\overrightarrow{A}\times \overrightarrow{B}\right|/AB\right]=11.9°}$.

Explanation of Solution

Take the cross product of vector $A$, and $B$.

  $\begin{array}{c}\overrightarrow{A}\times \overrightarrow{B}=\left|\begin{array}{l}\widehat{i}\widehat{j}\widehat{k}\\ -3.007.00-4.00\\ 6.00-10.09.00\end{array}\right|\\ =23.0\widehat{i}+3.00\widehat{j}-12.0\widehat{k}\end{array}$

Consider the magnitude of $\overrightarrow{A}\times \overrightarrow{B}$.

  $\begin{array}{c}\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{{\left(23.0\right)}^2+{\left(3.00\right)}^2+{\left(-12.0\right)}^2}\\ =26.1\end{array}$

Conclusion:

Substitute, $26.1$ for $\left|\overrightarrow{A}\times \overrightarrow{B}\right|$, $127$ for $AB$ in ${\sin }^{-1}\left[\left|\overrightarrow{A}\times \overrightarrow{B}\right|/AB\right]$.

  $\begin{array}{c}{\sin }^{-1}\left[\left|\overrightarrow{A}\times \overrightarrow{B}\right|/AB\right]={\sin }^{-1}\left(\frac{26.1}{127}\right)\\ =11.9°\end{array}$

Therefore, the $\underset{\_}{{\sin }^{-1}\left[\left|\overrightarrow{A}\times \overrightarrow{B}\right|/AB\right]=11.9°}$.

(c)

To determine

Out of the above two methods which gives the angle between vectors.

Answer to Problem 24P

The first method gives the angle between vectors.

Explanation of Solution

It sis known that $\sin \left(180-\theta \right)$ is equal to $\sin \theta$, but $\cos \left(180-\theta \right)$ is equal to $-\cos \theta$. The first one $\sin \left(180-\theta \right)$ makes no difference between $180-\theta$ and $\theta$.

Hence the method in part (a) gives the angle between vectors.

Conclusion:

Therefore, the first method gives the angle between vectors.